我正在使用 DRF 提供的 DefaultRouter,因为我需要一个根 api View 。但是,该 View 上的项目没有任何逻辑顺序。我查看了源代码,发现每个条目都被放入字典中(本质上没有排序)。
class DefaultRouter(SimpleRouter):
"""
The default router extends the SimpleRouter, but also adds in a default
API root view, and adds format suffix patterns to the URLs.
"""
include_root_view = True
include_format_suffixes = True
root_view_name = 'api-root'
def get_api_root_view(self):
"""
Return a view to use as the API root.
"""
api_root_dict = {}
list_name = self.routes[0].name
for prefix, viewset, basename in self.registry:
api_root_dict[prefix] = list_name.format(basename=basename)
class APIRoot(views.APIView):
_ignore_model_permissions = True
def get(self, request, format=None):
ret = {}
for key, url_name in api_root_dict.items():
ret[key] = reverse(url_name, request=request, format=format)
return Response(ret)
return APIRoot.as_view()
我想按字母顺序对根 api View 上的项目进行排序,并且可以通过修改源轻松实现。但我想知道,你们有没有想过在不修改源代码的情况下订购根 api 项目的解决方案?
最佳答案
根据您的建议和 Denis Cornehi 答案中的第一点,这是 DefaultRouter 的扩展,它按 base_names 对 url 进行排序:
# myapp/routers.py
from rest_framework import routers
from rest_framework import views
from rest_framework.response import Response
from rest_framework.reverse import reverse
import operator
import collections
class OrderedDefaultRouter(routers.DefaultRouter):
def get_api_root_view(self):
"""
Return a view to use as the API root but do it with ordered links.
"""
api_root_dict = {}
list_name = self.routes[0].name
for prefix, viewset, basename in self.registry:
api_root_dict[prefix] = list_name.format(basename=basename)
class APIRoot(views.APIView):
_ignore_model_permissions = True
def get(self, request, format=None):
ret = {}
for key, url_name in api_root_dict.items():
ret[key] = reverse(url_name, request=request, format=format)
sorted_ret = collections.OrderedDict(sorted(ret.items(), key=operator.itemgetter(0)))
return Response(sorted_ret)
return APIRoot.as_view()
关于django - 在 DefaultRouter 的根 api View 上订购项目,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24311893/